# How do you solve the system by graphing and what is the ordered pair for x - 2y = 0 and 5x+2y=24?

Aug 20, 2017

See a solution process below:

#### Explanation:

For each equation we can find two solutions for the equation, plot the points and draw a line through the two points to graph the line.

Equation 1

For $x = 0$

$0 - 2 y = 0$

$- 2 y = 0$

$\frac{- 2 y}{\textcolor{red}{- 2}} = \frac{0}{\textcolor{red}{- 2}}$

$y = 0$ or $\left(0 , 0\right)$

For $x = 4$

$4 - 2 y = 0$

$- \textcolor{red}{4} + 4 - 2 y = - \textcolor{red}{4} + 0$

$0 - 2 y = - 4$

$- 2 y = - 4$

$\frac{- 2 y}{\textcolor{red}{- 2}} = \frac{- 4}{\textcolor{red}{- 2}}$

$y = 2$ or $\left(4 , 2\right)$

graph{(x^2+y^2-0.25)((x-4)^2+(y-2)^2-0.25)(x-2y)=0 [-30, 30, -15, 15]}

Equation 2

For $x = 0$

$\left(5 \cdot 0\right) + 2 y = 24$

$0 + 2 y = 24$

$2 y = 24$

$\frac{2 y}{\textcolor{red}{2}} = \frac{24}{\textcolor{red}{2}}$

$y = 12$ or $\left(0 , 12\right)$

For $x = 2$

$\left(5 \cdot 2\right) + 2 y = 24$

$10 + 2 y = 24$

$- \textcolor{red}{10} + 10 + 2 y = - \textcolor{red}{10} + 24$

$0 + 2 y = 14$

$2 y = 14$

$\frac{2 y}{\textcolor{red}{2}} = \frac{14}{\textcolor{red}{2}}$

$y = 7$ or $\left(2 , 7\right)$

graph{(x^2+(y-12)^2-0.25)((x-2)^2+(y-7)^2-0.25)(5x+2y-24)(x-2y)=0 [-30, 30, -15, 15]}

Solution

We can see below the lines intersect at: (4, 2)

graph{(5x+2y-24)(x-2y)=0 [-12, 12, -6, 6]}